I need to find the derivative of this function. I think I managed to do this, but the simplification is messy. Is there a good way to keep things organized while I go through all of the needed differentiation rules? It was hard to stay focused in even writing this in latex.
$y = \frac{\sqrt[3]{2x}\sqrt{x+5}}{(4x-1)^9}$
numerator(u)/denominator(v)
$u = \sqrt[3]{2x}\sqrt{x+5}$
$v = (4x-1)^9$derivative of numerator(u')/denominator(v')
$u' =\frac{(2x)^{\frac{1}{3}}(x+5)^{-\frac{1}{2}}}{2} + \frac{(x+5)^{\frac{1}{2}}(2x)^{-\frac{2}{3}}}{3}$ product rule, chain rule
$v' = 36(4x-1)^8$ chain rule
quotient rule $\frac{vu' - uv'}{v^2}$
$= \frac{\bigg(\frac{(2x)^{\frac{1}{3}}(x+5)^{-\frac{1}{2}}}{2} + \frac{(x+5)^{\frac{1}{2}}(2x)^{-\frac{2}{3}}}{3} \bigg) - 36\sqrt[3]{2x}\sqrt{x+5}\ (4x-1)^8}{(4x-1)^9}$
$= \frac{\bigg(\frac{\sqrt[3]{2x}}{2\sqrt{x+5}} + \frac{\sqrt{x+5}}{6\sqrt[3]{2x}} \bigg) - 36\sqrt[3]{2x}\sqrt{x+5}\ (4x-1)^8}{(4x-1)^9}$
Suffice to say this is a mess, does a less-tedious way exist to find the derivative?
The classical trick is to consider \begin{align} \ln y = \frac{1}{3}\ln(2x)+\frac{1}{2}\ln(x+5)-9\ln(4x-1) \end{align} then differentiate to get \begin{align} \frac{y'}{y} =&\ \frac{1}{3x}+\frac{1}{2(x+5)}-\frac{36}{4x-1}\\ =&\ -\frac{196x^2+1045x+10}{6x(x+5)(4x-1)}. \end{align} Hence it follows \begin{align} y' =&\ -y\frac{196x^2+1045x+10}{6x(x+5)(4x-1)} = - \frac{\sqrt[3]{2x}\sqrt{x+5}}{(4x-1)^9}\frac{196x^2+1045x+10}{6x(x+5)(4x-1)} \\ =&\ - \frac{196x^2+1045x+10}{3\cdot 2^{2/3} x^{2/3}\sqrt{x+5}(4x-1)^{10}} \end{align}
Remark: In the grand scheme of things, it doesn't save much time in terms of getting a simplified expression. However, it does have some organizational benefits.
Remark $2$: As mentioned by egreg, we can also write \begin{align} \ln |y| = \ln\left|\frac{\sqrt[3]{2x}\sqrt{x+5}}{(4x-1)^9}\right| = \frac{1}{3}\ln|2x|+\frac{1}{2}\ln|x+5|-9\ln|4x-1| \end{align} and use the fact that \begin{align} \frac{d}{dx}\ln|f(x)| = \frac{f'(x)}{f(x)}. \end{align}